A 20 g bullet is accelerated in a rifle barrel
45 cm long to a speed of 354 m/s.
Use the work-energy theorem to find the
average force exerted on the bullet while it is
being accelerated.
Answer in units of N.
Well, let me calculate that for you using the work-energy theorem. But first, let me tell you a joke to lighten the mood.
Why don't scientists trust atoms?
Because they make up everything!
Now, back to the calculation. According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. The formula for work is given by:
Work = Force × Distance
In this case, the distance is the length of the rifle barrel, which is 45 cm (or 0.45 m), and the change in kinetic energy can be calculated using the formula:
Change in Kinetic Energy = (1/2) × mass × (final velocity^2 - initial velocity^2)
So, let's plug in the values and calculate:
Change in Kinetic Energy = (1/2) × 0.02 kg × (354 m/s)^2 - 0
Now, we can equate the work done to the change in kinetic energy:
Force × Distance = Change in Kinetic Energy
Force × 0.45 m = (1/2) × 0.02 kg × (354 m/s)^2
Simplifying, we get:
Force = [(1/2) × 0.02 kg × (354 m/s)^2] / 0.45 m
Calculating that, we find:
Force ≈ 69,942.7 N
So, the average force exerted on the bullet while it is being accelerated is approximately 69,942.7 N.
And remember, don't shoot the messenger... or the clown!
To find the average force exerted on the bullet while it is being accelerated, we can use the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy.
The work done on the bullet is equal to the change in its kinetic energy. The initial kinetic energy of the bullet is zero since it starts from rest, and the final kinetic energy is given by:
Kf = (1/2)mv^2
Where m is the mass of the bullet and v is its final velocity.
We know that the mass of the bullet is 20 grams, or 0.02 kg, and the final velocity is 354 m/s.
Kf = (1/2)(0.02 kg)(354 m/s)^2
= (1/2)(0.02 kg)(125316 m^2/s^2)
= 1253.16 J
The work done on the bullet is equal to the change in kinetic energy:
W = Kf - Ki
Since the initial kinetic energy is zero, the work done is equal to the final kinetic energy:
W = 1253.16 J
The average force exerted on the bullet is equal to the work done divided by the distance traveled. The distance traveled by the bullet is given as 45 cm, or 0.45 m.
Average force = W / d
= 1253.16 J / 0.45 m
= 2784.8 N
Therefore, the average force exerted on the bullet while it is being accelerated is 2784.8 N.
To find the average force exerted on the bullet while it is being accelerated, we can use the work-energy theorem. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
First, we need to calculate the initial and final kinetic energies of the bullet. The kinetic energy of an object is given by the equation:
KE = 1/2 * m * v^2
Where KE is the kinetic energy, m is the mass of the object, and v is its velocity.
Given:
Mass of the bullet, m = 20 g = 0.02 kg
Initial velocity, u = 0 m/s (since the bullet starts from rest)
Final velocity, v = 354 m/s
Initial kinetic energy, KE_initial = 1/2 * m * u^2 = 0
Final kinetic energy, KE_final = 1/2 * m * v^2
Now, using the work-energy theorem, we can calculate the work done on the bullet:
Work = KE_final - KE_initial
Substituting the values, we get:
Work = 1/2 * m * v^2 - 0
The average force exerted on the bullet is equal to the work done divided by the displacement. In this case, the displacement is the length of the barrel, which is given as 45 cm = 0.45 m.
Average force, F = Work / Displacement
Substituting the values, we have:
F = (1/2 * m * v^2 - 0) / 0.45
Finally, calculating the expression:
F = (1/2 * 0.02 kg * (354 m/s)^2) / 0.45
Simplifying the expression:
F = (0.01 kg * 125316 m^2/s^2) / 0.45
F = 27.87 N
Therefore, the average force exerted on the bullet while it is being accelerated is approximately 27.87 N.
Ke = (1/2) m v^2
= (1/2)(.020)(354)^2
= F * d = 0.45 F